An offset between the value of the controlled variable and its set-point remains

We observe that by increasing (in absolute value) 𝐾𝑐, the final steady state value of ℎ′(𝑡)decreases, i.e. it gets closer to the set-point of 0. Correspondingly the error 𝜀decreases (in absolute value). However, an offset between the value of the controlled variable and its set-point remains. This is as expected for a P controller. Note that in this example the values of 𝐾𝑐have to be negative. This is because when the level increases, the error 𝜀(𝑡)is negative and we want an increase in the 𝑐′(𝑡)signal, i.e. an increase in theoutlet flow rate. Therefore, 𝐾𝑐has to be negative to convert a negative 𝜀(𝑡)into a positive 𝑐′(𝑡).d) A change in the disturbance is considered, so the closed loop response equation for the regulator problem should be used. Furthermore, it is given that the disturbance undergoes a step change of size 0.001 m3/s, so its Laplace transform is 𝐹𝑖′(𝑠)=0.001𝑠. We have a P controller only, so the closed loop response equation becomes:-1.5-1-0.500.511.502004006008001000h'(t), ε(t) [m]time [s]h (no control)ε(no control)-1.5-1-0.500.511.502004006008001000h'(t), ε(t) [m]time [s]h (no control)ε(no control)h (Kc=-0.3)h (Kc=-1)ε(Kc=-0.3)ε(Kc=-1)
ℎ′(𝑠)=𝐺𝑑(𝑠)1+𝐺𝑓(𝑠)∙𝐺𝑐(𝑠)∙𝐺𝑝(𝑠)⋅𝐺𝑚(𝑠)⋅𝐹𝑖′(𝑠)=1s1+0.01∙𝐾𝑐∙(−1s)⋅1⋅0.001𝑠=0.001s(1+0.01∙𝐾𝑐∙(−1s))𝑠=0.001(𝑠−0.01∙𝐾𝑐)𝑠The final state steady value of ℎ′(𝑡)predicted by the above equation can be found in a variety of ways. Using e.g. the Final Value Theorem it can be found:lim𝑡→∞(ℎ′(𝑡))=lim𝑠→0(𝑠ℎ′(𝑠))=lim𝑠→0(𝑠0.001(𝑠−0.01∙𝐾𝑐)𝑠)=lim𝑠→0(0.001(𝑠−0.01∙𝐾𝑐))=−0.1𝐾𝑐The offset can be calculated from: 𝑜𝑓𝑓𝑠𝑒𝑡=𝑛𝑒𝑤_𝑠𝑝−𝑓𝑖𝑛𝑎𝑙_𝑣𝑎𝑙𝑢𝑒=0−(−0.1𝐾𝑐)=0.1𝐾𝑐. For the 𝐾𝑐values studied we hence have:𝐾𝑐𝑜𝑓𝑓𝑠𝑒𝑡-0.3-0.333-1-0.1It can be verified from the plots obtained above that the calculated offsets agree with the simulated ones. It can also be seen that as 𝐾𝑐increases the offset decreases, as expected from the analysis of P controllers.e) The comparison of P and PI control actions should look as follows:-0.15-0.1-0.0500.050.10.1502004006008001000h'(t), ε(t) [m]time [s]h (P, Kc=-1)h (PI, Kc=-1, τΙ=20)ε(P, Kc=-1εh (PI, Kc=-1, τΙ=20)
This comparison is according to the expected differences between the P and PI control actions. The PI action removes the offset and the final value of the error is0.This is as expected due to thedefinition of the PI control action:𝑐′(𝑡)=𝐾𝑐𝜀(𝑡)+𝐾𝑐𝜏𝐼∫𝜀(𝑡)𝑑𝑡𝑡0The only way to obtain a 𝑐′(𝑡)value at the new steady state which is constant and different from the initial value (assumed to be 0) is to have an error 𝜀(𝑡)=0at the new steady state. When 𝜀(𝑡)=0, the integralterm in the PI control action is constant and ≠0.On the other hand, for a P controller the only way to have a 𝑐′(𝑡)value at the new steady statedifferent than 0 is to have an error different than 0.Itcan be observed that the new steady state value of the controller signal 𝑐′(𝑡)is the same for the Pand PI controller. This is as expected, since the value of 𝑐′(𝑡)determines (via the transfer function ofthe valve) the value of the manipulated variable𝐹𝑜′(𝑡), which should be the same for any controllertype, and should be the same as the input feed 𝐹𝑖′(𝑡)(the liquid level should be constant at the newsteady state).However, the PI controller brings the disadvantage of the oscillatory response, which is an outcome of the increase in the order of the closed-loop system in comparison to the open loop system by the addition of the I action. Note that when the error reaches 0 for the first time, the value of the controller signal 𝑐′(𝑡)is at itsmaximum and not at the final steady state value. Therefore, the value of the integral has to decreaseand, to do so, the error has to change sign, i.e. the response has to be oscillatory